$G$-invariant morphism and coarse moduli spaces

Let $$G$$ be an algebraic group acting on $$X$$ (a finite type scheme on $$k$$).

A $$G$$-invariant $$k$$-morphism $$f : X \rightarrow S$$ is a map such that the following commute: $$\require{AMScd}$$ $$\begin{CD} G \times_k X @>\rho>> X\\ @V \pi_2 V V @VV f V\\ X @>>f> S \end{CD}$$

Where $$\rho$$ is the action map and $$\pi_2$$ is the projection on the second coordinate.

Every $$k$$-point $$g \in G(k)$$ induces a map $$\phi_g : X \rightarrow X$$ given by the composition: $$\require{AMScd}$$ $$\begin{CD} X = \mathop{Spec} k \times_k X @>{(g, Id)}>> G \times_k X @>{\rho}>> X \end{CD}$$ We now define a second kind of invariance: $$f : X \rightarrow S$$ is invariant if $$f= f \circ \phi_g$$ for all $$g \in G(k)$$.

Obviously, the first definition implies the second one. Is it true the inverse implication?

In my case $$k$$ is algebraically closed, but I don't know if I need more assumptions. If yes, do you know some counterexamples? It is true if I suppose $$X$$ and $$G$$ reduced, but I hope I can avoid this.

Furtermore: Let's consider the map induced on $$k$$-points: $$\rho(k) : G(k) \times X(k) \rightarrow X(k)$$. Let $$k$$ be algebraically closed and let $$X$$, $$G$$ be reduced. It's well known that $$\rho(k)$$ determines $$\rho$$. It is true if I do not assume $$G$$ to be reduced?

My real question: My question comes from coarse moduli space, in particular from proposition 3.35 of these notes. It isn't clear to me why $$\eta_S(\mathcal{F})$$ is $$G$$-invariant (with reference to pdf notations). The proposition doesn't request that $$X$$ and/or $$G$$ are reduced, but it seems that the proof assumes it.

An algebraic group is a group scheme of finite type over $$k$$. Every scheme is intended to be of finite type over $$k$$.

This is not true without the assumption that $$G$$ is reduced. Here is a counterexample.

Fix a prime number $$p$$ and any field $$k$$ of characteristic $$p$$. We define the nonreduced algebraic subgroup $$\alpha_p \subset \mathbb{G}_a$$ by $$\alpha_p = Spec(k[t]/(t^p))$$ (here $$\mathbb{G}_a$$ denotes the additive group over $$k$$). Then $$\alpha_p$$ contains a single $$k$$-point-- the identity $$0$$ of the group.

We set $$X = \mathbb{G}_a$$. The group $$\alpha_p$$ acts on $$X$$ by addition (include $$\alpha_p$$ into $$\mathbb{G}_a$$ and then use the group structure).

The identity morphism $$id: X \to X$$ is not $$\alpha_p$$-invariant (the scheme-theoretic image of $$\alpha_p \times 0 \subset \alpha_p \times X$$ is just $$0\subset X$$ under the projection and $$\alpha_p \subset X$$ under the action). However, it is clearly invariant under the unique $$k$$-point of $$\alpha_p$$ (the identity of the group).

If $$G$$ is reduced and $$S$$ is separated, then I think that you should be fine (under your assumption that $$k$$ is algebraically closed).

Edited:

For the last part, you can argue as follows. Consider the morphism $$\bigsqcup_{p \in G(k)} p \to G$$ By the Nullstellensatz, this has schematic image $$G$$. Using an argument similar to the end of the proof of Prop 3.2.4 (ii) in http://virtualmath1.stanford.edu/~conrad/249BW16Page/handouts/alggroups.pdf, you can show by arguing affine locally that the product morphism $$\bigsqcup_{p \in G(k)} p \times X \to G \times X$$ has schematic image $$G \times X$$. (Notice that the original morphism is not quasicompact! so it is not automatic that the schematic image commutes with flat base-change, but in this case it does. This is true when your base is a field because any ring over $$k$$ is a free $$k$$-module).

Now, if $$S$$ is separated, the locus of $$G \times X$$ where the projection agrees with the action is a closed subscheme $$Z \subset G \times X$$. By assumption, we have a factorization $$\bigsqcup_{p \in G(k)} p \times X \to Z \to G \times X$$ Therefore, by the scheme theoretic density we must have $$Z = G \times X$$.

• Thanks for your answer @afh. I'm happy to know that we can avoid assuming $X$ is reduced. Do you have any references for the last sentence you wrote? The notes I linked above still confuse me. Do you know if it's enough to assume that $G$ is reduced to prove proposition 3.35? Aug 16, 2021 at 12:25
• Hello @merlino. I added a small explanation to my answer. For your other question, it seems to me that in Prop. 3.35 there are some implicit assumptions the author had in mind. Otherwise the proposition cannot be true as stated. The first problem is in forming the G-equivariant map as in your question (the forward direction). For that you probably want to impose some condition on $S$ (e.g. reducedness).
– afh
Aug 17, 2021 at 3:24
• The other direction depends on your "moduli functor" $\mathcal{M}$. E.g. you can take $\mathcal{M}$ to be represented by a scheme and take $S$ to the trivial square-zero thickening $\mathcal{M}[\epsilon]$ with trivial action. Then the construction of the inverse does not go through.
– afh
Aug 17, 2021 at 3:24
• Thanks very much for your edit! Now it's clear to me why we need $S$ separated. I think that in this case is automatic that schematic image commute with base change. According to prop 0.4 in these notes, we need that the schematic image of $\bigsqcup_{p \in G(k)} p \to G$ is quasi-compact (that's true because it is $G$) and that $G$ is quasi-separated (that's true because it is of finite presentation). Aug 18, 2021 at 15:18
• You use $k$ algebraically closed only to show that $k$ points are dense. So, if we suppose $G$ reduced and $\mathop{char} k = 0$, we have that this result is true in general. Yes, probably the assumption that the author has in mind is $S$ reduced, but she doesn't mention it. I'm looking for a more general statement of this kind that does not require $S$ reduced. It's not clear to me even if it is a good purpose. Aug 18, 2021 at 15:19